# [SOLVED] Semi-direct product

• May 22nd 2008, 07:14 AM
topsquark
[SOLVED] Semi-direct product
As the work involved with helping me with this question might be rather extreme, I'll be satisfied with a pointer in the right direction. :)

I am trying to construct a group G as the semi-direct product of two groups H ($\displaystyle \{ e, a, b, c \}$) and F ( $\displaystyle \{ \epsilon, x \}$.) The multiplication tables for each are shown below in the attachment.

In order to do this I need to define the group elements for G. They are simply the direct product (taken in the sense of sets) of the groups F and H: $\displaystyle \{ (f, h)| \forall f \in F, h \in H \}$

The multiplication of two elements of G is defined as
$\displaystyle (f_1, h_1) ~ (f_2, h_2) = (f_1f_2, (h_1(f_2T))h_2)$
where $\displaystyle T \in Hom(F, Aut(H))$
(I don't know what is standard notation here. Wikipedia would write it as
$\displaystyle (f_1, h_1) ~ (f_2, h_2) = (f_1f_2, (T_{f_2}(h_1)h_2)$)

I have chosen
$\displaystyle T:~F \to Aut(H)$ as $\displaystyle T:~ \epsilon \to B, x \to A$ where A and B are two automorphisms on H:
$\displaystyle A:~H \to H$ as $\displaystyle A:~e \to a, a \to b, b \to c, c \to e$

$\displaystyle B:~H \to H$ as $\displaystyle B:~e \to c, a \to e, b \to a, c \to b$

The development of the multiplication table for G should be relatively simple: just apply the definition. But I am coming up with two strange facts before I even get very far.

First, $\displaystyle (\epsilon, e)$ is not the identity in G.
$\displaystyle (\epsilon, e)~(\epsilon, e) = (\epsilon \epsilon, (e( \epsilon T))e) = (\epsilon, (eB)e) = (\epsilon, c)$
I would have expected $\displaystyle (\epsilon, e)$ to be the identity for G.

Second, what I appeared to find as the identity, $\displaystyle (\epsilon, a)$, is only the identity for part of the group. We have
$\displaystyle (\epsilon , a)~(\epsilon , h) = (\epsilon, h)~\forall h \in H$
but
$\displaystyle (\epsilon, a)~(x, h) \neq (x, h)~\text{for any } h \in H$

So $\displaystyle (\epsilon, a)$ cannot be the identity either. In fact, though I admittedly haven't gone through the whole multiplication table, it would seem that G does not have an identity!

Obviously I'm doing something wrong. As I've never worked with semi-direct products before I'm guessing I somehow defined my function T wrong? Or is it something else?

Thanks!
-Dan
• May 22nd 2008, 08:14 AM
topsquark
Okay, I've finished the multiplication table. Some disturbing features have cropped up. There is a right inverse, but no left inverse and the "group" is not associative.

I've made a mistake somewhere, but I've gone over the calculations in detail and can't find it. Perhaps my definition of the group multiplication is wrong? (That would be far from the first typo I've seen in this text.)

-Dan
• May 23rd 2008, 12:37 AM
Opalg
Quote:

Originally Posted by topsquark
I am trying to construct a group G as the semi-direct product of two groups H ($\displaystyle \{ e, a, b, c \}$) and F ( $\displaystyle \{ \epsilon, x \}$.) The multiplication tables for each are shown below in the attachment.

In order to do this I need to define the group elements for G. They are simply the direct product (taken in the sense of sets) of the groups F and H: $\displaystyle \{ (f, h)| \forall f \in F, h \in H \}$

The multiplication of two elements of G is defined as
$\displaystyle (f_1, h_1) ~ (f_2, h_2) = (f_1f_2, (h_1(f_2T))h_2)$
where $\displaystyle T \in Hom(F, Aut(H))$
(I don't know what is standard notation here. Wikipedia would write it as
$\displaystyle (f_1, h_1) ~ (f_2, h_2) = (f_1f_2, (T_{f_2}(h_1)h_2)$)

First comment: Algebraists tend to write operators on the right, so fT means an element f acted on by a transformation T. Analysts write it the other way round, T(f), thinking of T as a function that acts on the element f. I'm an analyst, so I'll use the Wikipedia notation.

Quote:

Originally Posted by topsquark
I have chosen
$\displaystyle T:~F \to Aut(H)$ as $\displaystyle T:~ \epsilon \to B, x \to A$ where A and B are two automorphisms on H:
$\displaystyle A:~H \to H$ as $\displaystyle A:~e \to a, a \to b, b \to c, c \to e$

$\displaystyle B:~H \to H$ as $\displaystyle B:~e \to c, a \to e, b \to a, c \to b$

Here's where the trouble starts! An automorphism is a mapping from a group to itself that preserves the group structure. In particular, it must send the identity element to itself. Neither of the maps A, B does this, so these maps are not automorphisms. In fact, H is the cyclic group with four elements; e is the identity element, b has order 2, and the other elements a and c have order 4. There are only two automorphisms of H. One of them is the identity mapping I (that sends each element to itself). The other automorphism, J say, exchanges a and c, and sends e and b to themselves: J(e)=e, J(a)=c, J(b)=b, J(c)=a.

Next, the mapping T is a homomorphism from F to Aut(H). Here again, that means that T has to preserve the group structure. So it has to take the identity element ε of F to the identity element I of the group Aut(H). There are two options for the action of T on x (the other element of F). We could have T(x)=I. But then the action of F on Aut(H) would be trivial, and the semidirect product would just be a direct product. To get a nontrivial semidirect product, we have to take T(x)=J.

Quote:

Originally Posted by topsquark
Obviously I'm doing something wrong. As I've never worked with semi-direct products before I'm guessing I somehow defined my function T wrong?

If you replace the maps B and A by I and J, and redefine T as above, then you'll find that the semidirect product $\displaystyle H\rtimes_TF$ is indeed a group, with identity element (e,ε). Geometrically, it represents the dihedral group D_8 of symmetries of a square. The subgroup H (or more accurately H×{ε}) is the group of rotations of the square, with F acting on H by conjugating with a reflection.
• May 23rd 2008, 05:00 AM
topsquark
Quote:

Originally Posted by Opalg
An automorphism is a mapping from a group to itself that preserves the group structure.

I do seem to remember that comment from somewhere. (It's probably right in front of me in my group theory book and I missed it when I was doing my brief review.) That makes much more sense, especially when I looked over my text's proof that this construction makes an associative group. I had already noted that the text was making some "assumptions" about the homomorphism, but had missed the point about the automorphism.

Thank you for your assistance. :)

-Dan